# Equivalence of definitions of open sets in metric space

Consider the following two equivalent definitions of an open set in a metric space $$(X,d)$$:

• Definition 1: A set $$U$$ in a metric space is open if and only if $$U$$ is an arbitrary union of open balls of elements in the metric space, or an open ball itself.

• Definition 2: A set $$U$$ in a metric space is open if and only if, for every element $$x\in X$$, there exists an $$\epsilon$$ such that an open ball $$B_{\epsilon}(x)$$ is contained within $$U$$.

Although Definition 2 clearly implies Definition 1, how does Definition 1 imply Definition 2?

• For a set $U$ described in definition 2, it can be represented as the union of all of the open ball witnin $U$ itself containing some $x \in U$ , and the index can run over all of $U$, which means every $x\in U$ belongs to one of the such ball. Commented Mar 23, 2022 at 12:42

Definition 1 implies 2 because of the observation that if $$B(x,r)$$ is an open ball and $$y\in B(x,r)$$ then for $$r'=r-d(x,y)$$ (which is positive) we have $$B(y,r')\subseteq B(x,r)$$. The last inclusion follows from the triangle inequality.